8. The Švarc-Milnor Lemma

Definition: A metric space is proper if closed balls of finite radius in are compact. The action of a group on a metric space is cocompact if is compact in the quotient topology.

The Švarc-Milnor Lemma: Let be a proper geodesic metric space. Let act cocompactly and properly discontinuously on . (Properly discontinuously means that for all compact .) Then is finitely generated and, for any , the map

is a quasi-isometry (where is equipped with the word metric).

Proof: We may assume that is infinite and is non-compact. Let be large enough that the -translates of cover . Set

Let . Let . We want to prove that:
(a) generates,
(b) ,

(c) , there exists such that

Note: and .

(c) is obvious.
(b-i) is also obvious.
To complete the proof we need to show (a) and (b-ii).

Assume . Let be such that

As , . Choose , such that and for each . Choose such that for each . Let , so . Now

So, . Therefore generates .

Also,

as required.

Corollary: If is a finite index subgroup of a finitely generated group then is quasi-isometric to .

Two groups and are commensurable if they have isomorphic subgroups of finite index. Clearly, if and are commensurable then they are quasi-isometric.

Example:.
Semidirect product is taken over the matrix This means that , but

Let with eigenvalues with . Let .

sits inside as a uniform lattice, meaning is a compact space.

Exercise 11: What is this quotient?

So, is a quasi-isomorphic to But, Bridson-Gersten showed that and are commensurable if and only if the corresponding eigenvalues have a common power.

Exercise 12: Let be the infinite regular valent tree. Prove that for all , is quasi-isometric to .